Hypothesis of Large Scale

7/28/2019 Hypothesis of Large Scale

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Piyoosh Bajoria 2

• Contents.

• 1. Elements of a statistical test

• 2. A Large-sample statistical test• 3. Testing a population mean

• 4. Testing a population proportion

• 5. Testing the difference between two populationmeans

• 6. Testing the difference between two population

proportions

• 7. Reporting results of statistical tests: p-Value



Varsha Varde 3

Mechanics of Hypothesis Testing

• Null Hypothesis :Ho:

What You Believe(Claim/Status quo)

• Alternative Hypothesis:Ha: The Opposite ( proveor disprove with samplestudy)

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Ha is less than type or left-tail test• 1. One-Sided Test of Hypothesis:

• < (Ha is less than type or left-tail test).• To see if a minimum standard is met

• Examples

• Contents of cold drink in a bottle

• Weight of rice in a pack

• Null hypothesis (H 0 ) : : µ = µ

0 Alternative hypothesis (Ha): : µ < µ0

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Ha is more than type or right -tail test

• One-Sided Test of Hypothesis:

• > (Ha is more than type or right -tail test).

• To see that maximum standards are not

exceeded.

• Examples

• Defectives In a Lot

• Accountant Claims that Hardly 1%

Account Statements Contain Error

• . Null hypothesis (H 0 ): p = p0

Alternative hypothesis (Ha): p > p0



Two-Sided Test of Hypothesis: • Two-Sided Test of Hypothesis:

• ≠ (Ha not equal to type)• Divergence in either direction is critical

• Examples

• Shirt Size of 42

• Size of Bolt & nuts

• Null hypothesis (H 0 ) : µ = µ0

Alternative hypothesis (Ha): µ ≠ µ0

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DEFINITIONS

• Type I error ≡{ reject H 0 |H 0 is true }

• Type II error ≡{ do not reject H 0 |H 0 isfalse}

• α = Prob{Type I error}• β= Prob{Type II error}

• Power of a statistical test:

Prob{reject H 0 |H 0 is false }= 1- β

Varsha Varde 77Piyoosh Bajoria

EXAMPLE



EXAMPLE• Example 1.

• H 0 : Innocent

• Ha: Guilty

• α = Prob{sending an innocent person to jail}

• β= Prob{letting a guilty person go free}

• Example 2.

• H 0 : New drug is not acceptable

• Ha: New drug is acceptable

• α = Prob{marketing a bad drug}• β= Prob{not marketing an acceptable drug}

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GENERAL PROCEDURE FOR HYPOTHESIS TESTING

• Formulate the null & alternative hypothesis

• Equality Sign Should Always Be In NullHypothesis

• Choose the appropriate sampling distribution

• Select the level of significance and hence thecritical values which specify the rejection and

acceptance region

• Compute the test statistics and compare it to

critical values

• Reject the Null Hypothesis if test statistics falls in

the rejection region .Otherwise accept it

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• Null hypothesis: H 0

• Alternative (research) hypothesis: H a• Test statistic:

• Rejection region : reject H 0 if .....

• Decision: either “Reject H 0 ” or “Do not reject H 0 ”

• Conclusion: At 100α % significance level there is

(in)sufficient statistical evidence to “ favour Ha” . • Comments:

• * H 0 represents the status-quo

• * Ha is the hypothesis that we want to provideevidence to justify. We show that Ha is true by

showing that H 0 is false, that is proof by contradiction.

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A general Large-Sample Statistical Test

• Parameter of interest: θ

• Sample data: n, ˆθ, σ ̂θ

• Other information: µ0 = target value,

α = Level of significance

• Test:Null hypothesis (H 0 ) : θ = θ 0

: Alternative hypothesis (Ha):

1) θ > θ 0 or

2) θ <θ 0 or

3) θ ≠θ 0

• Test statistic (TS): z =( ̂θ - θ 0 )/ σ ̂θ

• Critical value: either z α or z α /2

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A General Large-Sample Statistical Test

• Rejection region (RR) :

• 1) Reject H0 if z > z α • 2) Reject H0 if z < - z α

• 3) Reject H0 if z > z α /2 or z < -z α /2

Decision: 1) if observed value is in RR: “Reject H0”

• 2) if observed value is not in RR: “Do no reject H0”

• Conclusion: At 100α% significance level thereis (in)sufficient statistical evidence to…….. .

• Assumptions: Large sample + others (to be

specified in each case).

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T ti P l ti M



Testing a Population Mean

• Conclusion: At 100α%

significance level there is(in)suficient statistical evidence to

“ favour Ha” . • Assumptions:

• Large sample (n ≥30)

• Sample is randomly selected

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EXAMPLE• Example: It is claimed that weight loss in a new diet

program is at least 20 pounds during the first month.

Formulate &Test the appropriate hypothesis• Sample data : n = 36, x¯ = 21, s2 = 25, µ0 = 20, α = 0.05

• H0 : µ ≥20 ( µ is 20 or larger)

• Ha : µ < 20 (µ is less than 20)

• T.S. :z =(x - µ0 )/(s/√n)=21 – 20/ 5 /√36 = 1.2

• Critical value: z α = -1.645

• RR: Reject H0 if z < -1.645

• Decision: Do not reject H0 • Conclusion: At 5% significance level there is insufficient

statistical evidence to conclude that weight loss in a new

diet program exceeds 20 pounds per first month.

Pi oosh Ba oria

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Testing a Population Proportion

• Parameter of interest: p (unknown parameter)

• Sample data: n and x (or p = x/n)

• p0 = target value• α (significance level)

• Test:H0 : p = p0

• Ha: 1) p > p0 ; 2) p < p0 ; 3) p = p0

• T.S. :z =( p - p0 )/√p0 q0 /n

• Rejection region (RR) :

• 1) Reject H0 if z > z α

• 2) Reject H0 if z < - z α • 3) Reject H0 if z > z α /2 or z < -z α /2

• Decision: 1) if observed value is in RR: “Reject H0”

• 2) if observed value is not in RR: “Do no reject H0”

• Assumptions:1. Large sample (np≥ 5, nq≥ 5) 2. Sample isPiyoosh Bajoria



Example• Test the hypothesis that p > .10 for sample data:

• n = 200, x = 26.

• Solution. p = x/n = 26/ 200 = .13,

• H0 : p = .10 (p is not larger than .10)

• Ha : p > .10

• TS:z = (p - p0 )/√p0 q0 /n=.13 - .10/√(.10)(.90)/200 = 1.41• RR: reject H0 if z > 1.645

• Dec: Do not reject H 0

• Conclusion: At 5% significance level there is insufficient

statistical evidence to conclude that p > .10.• Exercise: Is the large sample assumption satisfied

here ?

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Comparing Two Population Means

• Parameter of interest: µ1 - µ2

• Sample data:

• Sample 1: n1, x¯ 1, s1

• Sample 2: n2 , x¯ 2 , s2

• Test:

• H 0 : µ1 - µ2 = D0

• Ha : 1)µ1 - µ2 > D0 ; 2) 1)µ1 - µ2 < D0 ;3) µ1 - µ2 = D0

• T.S. :z =( x¯ 1 - x¯ 2 ) - D0 /√σ 2 1/n1+ σ2

2/n2

• RR:1) Reject H0 if z > z α ;2) Reject H0 if z < -z α


• Assumptions:• 1. Large samples ( n1≥ 30; n2 ≥30)

• 2. Samples are randomly selected

• 3. Samples are independent

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• Refer to the weight loss example. Test the hypothesis

that weight loss in the two diet programs are different.

• 1. Sample 1 : n1 = 36, x¯ 1 = 21, s2 1 = 25 (old)

• 2. Sample 2 : n2 = 36, x¯ 2 = 18.5, s2 2 = 24 (new)

• D0 = 0, α = 0.05

• H0 : µ1 - µ2 = 0

• Ha : µ1

- µ2

≠ 0,

• T.S. :z =( x¯ 1 - x¯ 2 ) – 0/√σ 2 1/n1+ ó2

2/n2 = 2.14

• Critical value: z α /2 = 1.96

• RR: Reject H0 if z > 1.96 or z < -1.96

• Decision: Reject H0 • Conclusion: At 5% significance level there is sufficient

statistical evidence to conclude that weight loss in the

two diet programs are different.

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• Parameter of interest: p1 - p2

• Sample 1: n1, x 1, ˆp1 = x 1 /n1

• Sample 2: n2 , x 2 , ˆp2 = x 2 /n2

• p1 - p2 (unknown parameter)• Common estimate: ̂p =(x1 + x2)/(n1 + n2)

• Test:H 0 : p1 - p2 = 0

• Ha : 1) p1 - p2 > 0;2) p1 - p2 < 0;3) p1 - p2 = 0

• TEST STATISTICS:z =( ̂p1 - ˆp2) – 0/ ̂pˆq(1/n1 + 1/n2)

• RR:1) Reject H0 if z > z α

• 2) Reject H0 if z < -z α


• Assumptions:

• Large sample(n1 p1≥ 5, n1q1 ≥5, n2 p2 ≥5, n2 q2 ≥5) • Samples are randomly and independently selected

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Example • Test the hypothesis that p1 - p2 < 0 if i t is known that

the test stat ist ic is

• z = -1.91.

• Solution:

• H0 : p1 - p2 = 0

• Ha : p1 - p2 < 0

• TS: z = -1.91

• RR: reject H0 if z < -1.645

• Dec: reject H0 • Conclusion: At 5% significance level there is sufficient

statistical evidence to conclude

• that p1 - p2 < 0.

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Reporting Results of Statistical Tests: P-Value• Definition. The p-value for a test of a hypothesis is the smallest

value of α for which the null hypothesis is rejected, i.e. the statistical

results are significant.• The p-value is called the observed significance level

• Note: The p-value is the probability ( when H0 is true) of obtaining avalue of the test statistic as extreme or more extreme than the actual

sample value in support of Ha.

• Examples. Find the p-value in each case:

• (i) Upper tailed test:H0 : θ = θ 0 ;Ha : θ> θ 0 ;

• TS: z = 1.76 p-value = .0392

• (ii) Lower tailed test:H0 : θ = θ 0 ;Ha : θ < θ 0

• TS: z = -1.86 p-value = .0314• (iii) Two tailed test: H0 : θ = θ 0 ;Ha : θ≠ θ 0

• TS: z = 1.76 p-value = 2(.0392) = .0784

• Decision rule using p-value: (Important)

• Reject H0 for all α > p- valuePiyoosh Bajoria 22

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Hypothesis of Large Scale

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